Valid Palindrome

The Idea

A palindrome reads the same from both ends, so check it from both ends. Put one pointer at the start, one at the end, and walk them toward each other.

One mismatch is a definite no. Meeting in the middle is a definite yes — no reversed copy, no extra memory.

Real-World Example

Barcode and account-number checkers run the same shape of test: read from the front and the back at once and bail the instant the two disagree. Scanning the whole code first, then comparing, wastes the half of the work you already knew was doomed.

The Code

def is_palindrome(s):
    left, right = 0, len(s) - 1
    while left < right:
        if not s[left].isalnum():      # skip junk from the left
            left += 1
        elif not s[right].isalnum():   # ...and from the right
            right -= 1
        elif s[left].lower() != s[right].lower():
            return False               # one mismatch is enough
        else:
            left, right = left + 1, right - 1
    return True

print(is_palindrome("RaceCar!"))   # True
print(is_palindrome("byte"))       # False

def is_palindrome(s):
  left, right = 0, len(s) - 1
  while ___:
      if s[left] != s[right]:
          return False
      left, right = left + 1, right - 1
  return True

print(is_palindrome("abba"), is_palindrome("abca"))   # want True False